Chapter 16.2, Problem 16.131P

To determine

(a)

The acceleration of the centre of gravity.

Answer to Problem 16.131P

The acceleration of the centre of gravity is $R=37.8ft/s^2$.

Explanation of Solution

Given information:

Pole length, $L=20ft$

Weight, $W=100lb$

Angular velocity, $\omega =1rad/s$

Horizontal force, $P=120lb$

Pole mass, $m=\frac{W}{g}$

Here, $\begin{array}{l}W=100lb\\ g=32.2ft/s^2\end{array}$

$m=\frac{100}{32.2}=3.105slug$

Pole moment of inertia,

$I=\frac{m{L}^2}{12}$

$I=\frac{3.105\times {\left(20\right)}^2}{12}$

$I=103.5slug.f{t}^2$

Acceleration tangential component between G and C.

${(a_{G/C})}_t=\left(\left(L_{GC}\cos 10°\right)\alpha \right)\leftarrow +\left(\left(L_{GC}\sin 10°\right)\alpha \right)\downarrow$

Here, $L_{GC}$ = Distance between points G and C

$\alpha$ = Angular acceleration

Acceleration centripetal component between G and C.

${(a_{G/C})}_n=\left(\left(L_{GC}\sin 80°\right){\omega }^2\right)\downarrow +\left(\left(L_{GC}\cos 80°\right){\omega }^2\right)\to$

$\omega$ = Pole angular velocity

Point G acceleration,

$a_G=a_C+{(a_{G/C})}_t+{(a_{G/C})}_n$

$a_G=a_C+\left(\left(L_{GC}\cos 10°\right)\alpha \right)\leftarrow +\left(\left(L_{GC}\sin 10°\right)\alpha \right)\downarrow +\left(\left(L_{GC}\sin 80°\right){\omega }^2\right)\downarrow +\left(\left(L_{GC}\cos 80°\right){\omega }^2\right)\to$ …….(Equation A)

Force friction,

$F_f={\mu }_{k}N$

Here, ${\mu }_k=0.3$

$\therefore F_f=0.3N$

Moment at G,

$N{L}_{GC}\sin 10°-F_f{L}_{GC}\cos 10°+P{L}_{GC}\cos 10°-Ph=I\alpha$

Here,

$\begin{array}{l}l=20ft\\ P=120lb\\ {\mu }_k=0.3\\ I=103.52slug.f{t}^2\end{array}$

$\left\{N\left(\frac{l}{2}\right)\sin 10°-0.3N\left(\frac{l}{2}\right)\cos 10°+\left(120lb\right)\left(\frac{l}{2}\right)\cos 10°-\left(120lb\right)\left(6ft\right)\right\}=103.52\alpha$

$\left\{N\left(\frac{20ft}{2}\right)\sin 10°-0.3N\left(\frac{20ft}{2}\right)\cos 10°+\left(120lb\right)\left(\frac{20ft}{2}\right)\cos 10°-\left(120lb\right)\left(6ft\right)\right\}=103.52\alpha$ ….(Equation B)

$1.736N-9.848\left(0.3N\right)+1181.76-720=103.52\alpha$

$103.52\alpha +1.218N=460.769$

Vertical component of forces,

$N-mg=-\left(\frac{ml}{2}\alpha \right)\sin 10°-\left(\frac{ml}{2}{\omega }^2\right)\cos 10°$

Here,

$\begin{array}{l}m=3.1055slug\\ g=32.2ft/s^2\\ l=20lb\\ \omega =1rad/s\end{array}$

$N-\left(3.1055\times 32.2\right)=\left\{-\left(\frac{3.1055\times 20}{2}\alpha \right)\sin 10°-\left(\frac{3.1055\times 20}{2}{\left(1\right)}^2\right)\cos 10°\right\}$ …(Equation C)

$N-99.997=-5.392\alpha -30.58$

$N=69.41-5.392\alpha$

Normal force between pole and ground,

Substitute, $N=69.41-5.392\alpha$ in equation B,

$103.52\alpha +1.218\left(69.41-5.392\alpha \right)=460.769$

$\alpha =3.89rad/s^2$

So,

$N=69.41-5.392\alpha$

$N=69.41-5.392\left(3.89\right)$

$N=48.43lb$

Forces horizontal component,

$P-F_f=m{a}_C-\frac{ml}{2}\alpha \cos 10°+\frac{ml}{2}{\omega }^2\sin 10°$

Here,

$\begin{array}{l}{F}_f=0.3N\\ m=3.1055slug\\ P=120lb\\ l=20lb\\ \omega =1rad/s\\ N=48.43lb\\ \alpha =3.8909rad/s^2\end{array}$

$120-0.3N=\left\{\left(3.105\right)a_c-\frac{3.105\times 20}{2}\left(3.89\right)\cos 10°+\frac{3.105\times 20}{2}{\left(1\right)}^2\sin 10°\right\}$

$120-\left(0.3\times 48.433\right)=\left\{\left(3.105\right)a_c-\frac{3.105\times 20}{2}\left(3.89\right)\cos 10°+\frac{3.105\times 20}{2}{\left(1\right)}^2\sin 10°\right\}$

$105.47=3.105a_c-118.996+5.39$

$a_c=70.542ft/s^2$

Therefore,

$a_G=\left\{70.542+\left(\frac{l}{2}\left(3.89\right)\cos 10°\right)+\left(\frac{l}{2}\left(3.89\right)\sin 10°\right)+\left(\frac{l}{2}\left(1\right)\sin 80°\right)+\left(\frac{l}{2}\left(1\right)\cos 80°\right)\right\}$

$a_G=\left\{70.542+\left(\frac{20}{2}\left(3.89\right)\cos 10°\right)+\left(\frac{20}{2}\left(3.89\right)\sin 10°\right)+\left(\frac{20}{2}\left(1\right)\sin 80°\right)+\left(\frac{20}{2}\left(1\right)\cos 80°\right)\right\}$

$=\left\{\left(70.542-38.317+6.756+9.848+1.7364\right)\right\}$

$=\left(33.961ft/s^2\right)\to +\left(16.605ft/s^2\right)\downarrow$ ………(Equation D)

Acceleration at point G,

$a_G={\left(a_G\right)}_x+{\left(a_G\right)}_y$ ………….(Equation E)

Resultant acceleration,

$R=\sqrt{{\left(a_G\right)}^2{}_x+{\left(a_G\right)}^2{}_y}$

From equation D and E,

Substitute, $\begin{array}{l}{\left(a_G\right)}_x=33.961ft/s^2\\ {\left(a_G\right)}_y=16.6ft/s^2\end{array}$

$R=\sqrt{{\left(33.961\right)}^2+{\left(16.605\right)}^2}$

$R=37.8ft/s^2$

Resultant angle,

$\tan \beta =\frac{{\left(a_G\right)}_y}{{\left(a_G\right)}_x}$

$\tan \beta =\frac{16.605}{33.961}$

$\beta ={\tan }^{-1}\left(0.4889\right)$

$\beta =26.1°$

Conclusion:

The acceleration of the centre of gravity is $R=37.8ft/s^2$.

To determine

(b)

The normal force between pole and ground.

Answer to Problem 16.131P

The normal force between pole and ground is $N=48.43lb$.

Explanation of Solution

Given information:

Pole length, $L=20ft$

Weight, $W=100lb$

Angular velocity, $\omega =1rad/s$

Horizontal force, $P=120lb$

Pole mass, $m=\frac{W}{g}$

Here, $\begin{array}{l}W=100lb\\ g=32.2ft/s^2\end{array}$

$m=\frac{100}{32.2}=3.105slug$

Pole moment of inertia,

$I=\frac{m{L}^2}{12}$

$I=\frac{3.105\times {\left(20\right)}^2}{12}$

$I=103.5slug.f{t}^2$

Acceleration tangential component between G and C

${(a_{G/C})}_t=\left(\left(L_{GC}\cos 10°\right)\alpha \right)\leftarrow +\left(\left(L_{GC}\sin 10°\right)\alpha \right)\downarrow$

Here, $L_{GC}$ = Distance between points G and C

$\alpha$ = Angular acceleration

Acceleration centripetal component between G and C

${(a_{G/C})}_n=\left(\left(L_{GC}\sin 80°\right){\omega }^2\right)\downarrow +\left(\left(L_{GC}\cos 80°\right){\omega }^2\right)\to$

$\omega$ = Pole angular velocity

Point G acceleration,

$a_G=a_C+{(a_{G/C})}_t+{(a_{G/C})}_n$

$a_G=a_C+\left(\left(L_{GC}\cos 10°\right)\alpha \right)\leftarrow +\left(\left(L_{GC}\sin 10°\right)\alpha \right)\downarrow +\left(\left(L_{GC}\sin 80°\right){\omega }^2\right)\downarrow +\left(\left(L_{GC}\cos 80°\right){\omega }^2\right)\to$ …….(Equation A)

Force friction,

$F_f={\mu }_{k}N$

Here, ${\mu }_k=0.3$

$\therefore F_f=0.3N$

Moment at G,

$N{L}_{GC}\sin 10°-F_f{L}_{GC}\cos 10°+P{L}_{GC}\cos 10°-Ph=I\alpha$

Here,

$\begin{array}{l}l=20ft\\ P=120lb\\ {\mu }_k=0.3\\ I=103.52slug.f{t}^2\end{array}$

$\left\{N\left(\frac{l}{2}\right)\sin 10°-0.3N\left(\frac{l}{2}\right)\cos 10°+\left(120lb\right)\left(\frac{l}{2}\right)\cos 10°-\left(120lb\right)\left(6ft\right)\right\}=103.52\alpha$

$\left\{N\left(\frac{20ft}{2}\right)\sin 10°-0.3N\left(\frac{20ft}{2}\right)\cos 10°+\left(120lb\right)\left(\frac{20ft}{2}\right)\cos 10°-\left(120lb\right)\left(6ft\right)\right\}=103.52\alpha$ ….(Equation B)

$1.736N-9.848\left(0.3N\right)+1181.76-720=103.52\alpha$

$103.52\alpha +1.218N=460.769$

Vertical component of forces,

$N-mg=-\left(\frac{ml}{2}\alpha \right)\sin 10°-\left(\frac{ml}{2}{\omega }^2\right)\cos 10°$

Here,

$\begin{array}{l}m=3.1055slug\\ g=32.2ft/s^2\\ l=20lb\\ \omega =1rad/s\end{array}$

$N-\left(3.1055\times 32.2\right)=\left\{-\left(\frac{3.1055\times 20}{2}\alpha \right)\sin 10°-\left(\frac{3.1055\times 20}{2}{\left(1\right)}^2\right)\cos 10°\right\}$ …(Equation C)

$N-99.997=-5.392\alpha -30.58$

$N=69.41-5.392\alpha$

Normal force between pole and ground,

Substitute, $N=69.41-5.392\alpha$ in equation B,

$103.52\alpha +1.218\left(69.41-5.392\alpha \right)=460.769$

$\alpha =3.89rad/s^2$

So,

$N=69.41-5.392\alpha$

$N=69.41-5.392\left(3.89\right)$

$N=48.43lb$

Conclusion:

The normal force between pole and ground is $N=48.43lb$.